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 Post subject: Re: 1645 cal/cm^2, >100ft AFB at 34.5kV
PostPosted: Wed Sep 03, 2014 10:34 am 
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arcad wrote:
stevenal wrote:
The X/R refers to the source impedance, not the fault impedance (mostly resistive). Much less real prime mover power is needed to push the 40 kA through the mostly reactive source impedance.


I believe the discussion was about fault current and fault X/R, or the current and X/R at the point of fault. The source X/R is normally close to 0. The fault X/R factors in all the impedance (active and reactive) in the current path starting with the source all along to the point of fault. I don't see how you get more than 40kA fault current from 40kA active source. The picture below shows that current through an impedance connected between points A and B will be less than the source current:

Image


Would you mind sharing calculations proving what you say?


Fault impedance (or arc resistance) is the voltage at the fault location divided by the fault current, and is generally assumed be entirely resistive when not neglected altogether (bolted). Source impedance is the voltage behind the thevinin equivalent source divided by the current, and is highly reactive due to the nature of the generators, transformers and lines supplying the fault current. Pre-fault voltage is a good approximation of the thevenin voltage.

At 40kA and 34.5 kV I get a source impedance magnitude of 0.498 ohm. The angle of this impedance is atan(X/R) = 82.8 degrees. the real part of the source impedance is 0.498 cos(82.8) = 0.062 ohm. Real power to supply this bolted fault would be 40000^2*0.062*3= 296 MW. This is well short of the 2.5 GW claimed above.


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 Post subject: Re: 1645 cal/cm^2, >100ft AFB at 34.5kV
PostPosted: Wed Sep 03, 2014 8:43 pm 
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The model shows only a source and an arcing resistance, and does not consider a load nor time series analysis that give rise to the assymetrical current during a fault. The "source" includes both the upstream transofrmers, generators, and wiring, and downstrem loads.


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